Let us consider an incompressible generalized Burgers’ fluid bounded by a rigid plate at \(z=0\). The fluid occupies the porous half space \(z>0\). The z-axis is taken normal to the plate. Initially, the plate and fluid are at rest. After time \(t=0^{+}\), both the fluid and the plate perform solid body rotation with uniform angular velocity Ω about the z-axis. The fluid is electrically conducting under the action of a uniform magnetic field \(\mathbf{B}=\mathbf{B}_{0}+\mathbf{b}\), acting parallel to z-axis. Here \(\mathbf{B}_{0} \) is an applied magnetic field and b is an induced magnetic field. The electric field is assumed to be zero. The magnetic Reynolds number is taken small to neglect the effect of induced magnetic field. The flow in the fluid is induced due to the constantly accelerating plate. The governing flow equation is given by
$$ \biggl( 1+\lambda_{1}\frac{\partial}{\partial t}+ \lambda_{2}\frac {\partial ^{2}}{\partial t^{2}} \biggr) \biggl( \frac{\partial F}{\partial t}+2i\Omega F \biggr) =\left [ \begin{array}{@{}c@{}} \nu ( 1+\lambda_{3}\frac{\partial}{\partial t}+\lambda_{4}\frac{ \partial^{2}}{\partial t^{2}} ) \frac{\partial^{2}F}{\partial z^{2}} \\ -\frac{\sigma_{1}B_{0}^{2}}{\rho} ( 1+\lambda_{1}\frac{\partial }{\partial t}+\lambda_{2}\frac{\partial^{2}}{\partial t^{2}} ) F \\ -\frac{\nu\phi}{k} ( 1+\lambda_{3}\frac{\partial}{\partial t}+\lambda_{4}\frac{\partial^{2}}{\partial t^{2}} ) F\end{array} \right ] , $$
(1)
where \(F=u+iv \) is the complex velocity of the fluid and u and v are its components in x and y-directions, respectively, ρ is the fluid density, μ is the dynamic viscosity, ν is the kinematic viscosity, \(B_{0}\) is the magnitude of \(\mathbf{B}_{0}\), \(\sigma_{1}\) is the finite electrical conductivity of the fluid, ϕ (\(0<\phi <1\)) is the porosity and \(k>0\) is the permeability of the porous medium, \(\lambda_{1}\) and \(\lambda_{3}\) are relaxation and retardation times, whereas \(\lambda_{2}\) and \(\lambda_{4}\) are material constants having the dimensions as the square of time. The corresponding boundary and initial conditions are
$$\begin{aligned}& F ( 0,t ) =At; \qquad F ( \infty,t ) =0; \quad t>0, \end{aligned}$$
(2)
$$\begin{aligned}& F ( z,0 ) =\frac{\partial F ( z,0 ) }{\partial t}=\frac{\partial^{2}F ( z,0 ) }{\partial t^{2}}=0; \quad z>0, \end{aligned}$$
(3)
where A with dimension \(L/T^{2}\) is the constant acceleration of the plate in x-direction.
Introducing the following non-dimensional variables
$$ \xi=z \biggl( \frac{A}{\nu^{2}} \biggr)^{\frac{1}{3}}; \qquad \tau=t \biggl( \frac{A^{2}}{\nu} \biggr)^{\frac{1}{3}}; \qquad G= \frac{F}{ ( \nu A )^{\frac{1}{3}}}, $$
(4)
the dimensionless problem becomes
$$\begin{aligned}& \frac{\partial^{2}G ( \xi,\tau ) }{\partial\xi^{2}}+\gamma \frac{\partial^{3}G ( \xi,\tau ) }{\partial\tau\, \partial \xi ^{2}}+\eta\frac{\partial^{4}G ( \xi,\tau ) }{\partial\tau ^{2}\, \partial\xi^{2}}-\beta \frac{\partial^{3}G ( \xi,\tau ) }{\partial\tau^{3}} \\& \quad {}-a_{0}\frac{\partial^{2}G ( \xi,\tau ) }{\partial\tau^{2}} -b_{0} \frac{\partial G}{\partial\tau}-c_{0}G=0;\quad \xi,\tau>0, \end{aligned}$$
(5)
$$\begin{aligned}& G ( 0,\tau ) =\tau; \qquad G ( \infty,\tau ) =0; \quad \tau >0, \end{aligned}$$
(6)
$$\begin{aligned}& G ( \xi,0 ) =\frac{\partial G ( \xi,0 ) }{\partial \tau}=\frac{\partial^{2}G ( \xi,0 ) }{\partial\tau^{2}}=0;\quad \xi >0, \end{aligned}$$
(7)
where
$$\begin{aligned}& \alpha = \lambda_{1} \biggl( \frac{A^{2}}{\nu} \biggr) ^{\frac{1}{3}}, \qquad \beta=\lambda_{2} \biggl( \frac{A^{2}}{\nu} \biggr) ^{\frac{2}{3}}, \qquad \gamma=\lambda_{3} \biggl( \frac{A^{2}}{\nu} \biggr) ^{\frac{1}{3}}, \qquad \eta=\lambda_{4} \biggl( \frac{A^{2}}{\nu} \biggr) ^{\frac{2}{3}}, \\& \omega = \Omega \biggl( \frac{\nu}{A^{2}} \biggr) ^{\frac{1}{3}},\qquad M^{2}= \frac{\sigma_{1}B_{0}^{2}\nu^{\frac{1}{3}}}{\rho A^{\frac{2}{3}}},\qquad \frac{1 }{K}=\frac{\phi\mu}{k\rho} \biggl( \frac{\nu}{A^{2}} \biggr) ^{\frac {1}{3}}, \\& c_{0}=2i \omega+M^{2}+\frac{1}{K},\qquad a_{0} = \alpha+2i\omega\beta+M^{2}\beta+\frac{\eta}{K}; \qquad b_{0}=1+2i\omega\alpha+M^{2}\alpha+ \frac{\gamma}{K}. \end{aligned}$$
(8)
Here α and γ are the non-dimensional relaxation and retardation times, β and η are non-dimensional material constants, ω is the non-dimensional angular velocity, \(M^{2}\) is the non-dimensional magnetic parameter called Hartmann number and \(\frac{1}{K}\) is the non-dimensional porosity parameter, where the arbitrary constants \(a_{0}\), \(b_{0}\) and \(c_{0}\) are introduced for the sake of mathematical convenience.
The solution of Eq. (5) in view of Eqs. (6) and (7) in the transformed q-plane is given by
$$ \bar {G} ( \xi,q ) =\frac{1}{q^{2}}\exp \biggl[ -\sqrt { \frac{\beta q^{3}+a_{0}q^{2}+b_{0}q+c_{0}}{\eta q^{2}+\gamma q+1}}\xi \biggr] , $$
(9)
where
$$ \bar {G} ( \xi,q ) =\mathcal{L}\bigl\{ G ( \xi,\tau ) \bigr\} =\int _{0}^{\infty}\exp ( -q\tau ) G ( \xi,\tau ) \,\mathrm{d} \tau. $$
To find the Laplace inverse of Eq. (9), we use a similar procedure as in [22] and write \(\bar {G} ( \xi ,q ) \) in the product form as:
$$\begin{aligned}& \bar {G} ( \xi,q ) =\bar {G}_{1} ( q ) \bar {G}_{2} ( \xi,q ) , \end{aligned}$$
(10)
$$\begin{aligned}& \bar {G}_{1} ( q ) =\frac{1}{q^{2}}, \end{aligned}$$
(11)
$$\begin{aligned}& \bar {G}_{2} ( \xi,q ) =\exp \bigl( -\xi\sqrt{w ( q ) } \bigr) ; \qquad w ( q ) =\frac{\beta q^{3}+a_{0}q^{2}+b_{0}q+c_{0}}{\eta q^{2}+\gamma q+1}. \end{aligned}$$
(12)
Denoting \(G_{1} ( \tau ) =\mathcal{L}^{-1}\{\bar {G}_{1} ( q ) \} \) and \(G_{2} ( \xi,\tau ) =\mathcal{L}^{-1}\{\bar {G}_{2} ( \xi,q ) \}\), then by using the convolution theorem [23], we have
$$ G ( \xi,\tau ) = ( G_{1}\ast G_{2} ) ( \tau ) =\int_{0}^{\tau}G_{1} ( \tau-s ) G_{2} ( \xi ,s ) \,\mathrm{d}s. $$
(13)
The Laplace inverse of Eq. (11) yields
$$ G_{1} ( \tau ) =\tau. $$
(14)
To find \(G_{2} ( \xi,\tau ) =\mathcal{L}^{-1}\{\bar {G}_{2} ( \xi,q ) \}\), we use the inversion formula for compound functions [23]
$$ \mathcal{L}^{-1}\bigl\{ F \bigl[ w ( q ) \bigr] \bigr\} =\int _{0}^{\infty}f ( u ) g ( u,\tau ) \,\mathrm{d}u, $$
where \(f ( \tau ) =\mathcal{L}^{-1}\{F ( q ) \} \) and \(g ( u,\tau ) =\mathcal{L}^{-1}\{\exp ( -uw ( q ) ) \}\).
Choosing \(f ( \xi,q ) =\exp ( -\xi\sqrt{q} ) \), we have
$$\begin{aligned}& f ( \xi,\tau ) =\mathcal{L}^{-1}\bigl\{ \exp ( -\xi\sqrt{q} ) \bigr\} =\frac{\xi}{2\tau\sqrt{\pi\tau}}\exp \biggl( \frac{-\xi ^{2}}{4\tau} \biggr) ; \quad \xi>0, \end{aligned}$$
(15)
$$\begin{aligned}& G_{2} ( \xi,\tau ) =\mathcal{L}^{-1}\bigl\{ \bar {G}_{2} ( \xi ,q ) \bigr\} =\int_{0}^{\infty}f ( \xi,u ) g ( u,\tau ) \,\mathrm{d}u \\& \hphantom{G_{2} ( \xi,\tau )}=\frac{\xi}{2\sqrt{\pi}}\int_{0}^{\infty } \frac{1}{u\sqrt{u}}\exp \biggl( \frac{-\xi^{2}}{4u} \biggr) g ( u,\tau ) \, \mathrm{d}u. \end{aligned}$$
(16)
Now to find \(g ( u,\tau ) =\mathcal{L}^{-1}\{\exp ( -uw ( q ) ) \}\), we write \(w ( q ) \) in the following form:
$$ w ( q ) =\frac{\beta q^{3}+a_{0}q^{2}+b_{0}q+c_{0}}{\eta q^{2}+\gamma q+1}=\eta_{0}+\frac{\beta}{\eta}q+ \frac{\eta _{1}}{q-q_{1}}+\frac{\eta_{2}}{q-q_{2}}, $$
where
$$ \begin{aligned} &\eta_{0} =\frac{a_{0}}{\eta}- \frac{\beta\gamma}{\eta^{2}}; \qquad \eta _{1}=\frac{ ( A_{1}q_{1}+A_{2} ) \eta}{\sqrt{\gamma ^{2}-4\eta}};\qquad \eta_{2}=-\frac{ ( A_{1}q_{2}+A_{2} ) \eta}{\sqrt {\gamma^{2}-4\eta}}, \\ &A_{1} =b_{0}-\frac{\beta}{\eta}-\frac{a_{0}\gamma}{\eta}+ \frac {\beta\gamma^{2}}{\eta^{2}}; \qquad A_{2}=c_{0}-\frac{a_{0}}{\eta}+ \frac {\beta\gamma}{\eta^{2}}, \end{aligned} $$
(17)
and \(q_{1,2}=\frac{-\gamma\pm\sqrt{\gamma^{2}-4\eta}}{2\eta} \) are the roots of the equation \(\eta q^{2}+\gamma q+1=0\).
Thus
$$\begin{aligned} g ( u,\tau ) =&\mathcal{L}^{-1} \biggl\{ \exp ( -u\eta _{0} ) \exp \biggl( -\frac{u\beta}{\eta}q \biggr) \exp \biggl( - \frac{u\eta_{1}}{q-q_{1}} \biggr) \exp \biggl( -\frac{u\eta _{2}}{q-q_{2}} \biggr) \biggr\} \\ =&\exp ( -u\eta_{0} ) \mathcal{L}^{-1} \biggl\{ \exp \biggl( - \frac{u\beta}{\eta}q \biggr) \bigl[ 1-H_{1} ( q ) -H_{2} ( q ) +H_{1} ( q ) H_{2} ( q ) \bigr] \biggr\} , \end{aligned}$$
where \(H_{1} ( q ) =1-\exp ( -\frac{u\eta_{1}}{q-q_{1}} ) \) and \(H_{2} ( q ) =1-\exp ( -\frac{u\eta_{2}}{ q-q_{2}} ) \).
If we denote
$$\begin{aligned}& h_{1} ( \tau ) =\mathcal{L}^{-1}\bigl\{ H_{1} ( q ) \bigr\} =\sqrt{\frac{\eta_{1}u}{\tau}}\exp ( q_{1}\tau ) J_{1} ( 2\sqrt{\eta_{1}u\tau} ) , \\& h_{2} ( \tau ) =\mathcal{L}^{-1}\bigl\{ H_{2} ( q ) \bigr\} =\sqrt{\frac{\eta_{2}u}{\tau}}\exp ( q_{2}\tau ) J_{1} ( 2\sqrt{\eta_{2}u\tau} ) , \end{aligned}$$
where \(J_{1} ( \cdot ) \) is the Bessel function of the first kind of order one, then finally one has
$$\begin{aligned} g ( u,\tau ) =&e^{-u\eta_{0}}\delta \biggl( \tau- \frac {\beta u}{\eta} \biggr) -\sqrt{\eta_{1}u}\int_{0}^{\tau} \frac{\delta ( s-\frac{\beta u}{\eta} ) }{\sqrt{\tau-s}}\exp \bigl( q_{1} ( \tau -s ) -u\eta_{0} \bigr) \\ &{}\times J_{1} \bigl( 2\sqrt{\eta_{1}u ( \tau -s ) } \bigr) \, \mathrm{d}s \\ &{}-\sqrt{\eta_{2}u}\int_{0}^{\tau} \frac{\delta ( s-\frac{ \beta u}{\eta} ) }{\sqrt{\tau-s}}\exp \bigl( q_{2} ( \tau -s ) -u\eta_{0} \bigr) J_{1} \bigl( 2\sqrt{\eta_{2}u ( \tau -s ) } \bigr) \, \mathrm{d}s \\ &{}+u\sqrt{\eta_{1}\eta_{2}}\int_{0}^{\tau}\int_{0}^{\sigma} \frac{\delta ( \tau-s-\frac{\beta u}{\eta} ) }{\sqrt{\sigma ( s-\sigma ) }}\exp \bigl( q_{1}\sigma+q_{2} ( s-\sigma ) -u\eta _{0} \bigr) \\ &{}\times J_{1} ( 2\sqrt{\eta_{1}u \sigma} ) J_{1} \bigl( 2\sqrt{\eta_{2}u ( s-\sigma ) } \bigr) \,\mathrm{d}s \,\mathrm{d}\sigma, \end{aligned}$$
(18)
where \(\mathcal{L}^{-1}\{\exp ( -\alpha q ) \}=\delta ( \tau-\alpha ) \) and \(\delta ( \cdot ) \) indicates the Dirac delta function.
Invoking Eq. (18) into Eq. (16), we get
$$\begin{aligned} G_{2} ( \xi,\tau ) =&\frac{\xi}{2\sqrt{\pi}} \int _{0}^{\infty}\frac{\delta ( \tau-\frac{\beta u}{\eta } ) }{u\sqrt{u}}\exp \biggl( \frac{-\xi^{2}}{4u}-\eta_{0}u \biggr) \, \mathrm{d}u-\frac{\xi\sqrt{\eta_{1}}}{2\sqrt{\pi}} \int_{0}^{\tau }\int_{0}^{\infty} \frac{1}{u}\frac{\delta ( s-\frac{\beta u}{\eta} ) }{\sqrt{\tau-s}} \\ &{}\times\exp \biggl( \frac{-\xi^{2}}{4u}+q_{1} ( \tau-s ) - \eta_{0}u \biggr) J_{1} \bigl( 2\sqrt{\eta_{1}u ( \tau-s ) } \bigr) \,\mathrm{d}u \,\mathrm{d}s \\ &{}-\frac{\xi\sqrt{\eta_{2}}}{2\sqrt{\pi}}\int_{0}^{\tau }\int _{0}^{\infty}\frac{\delta ( s-\frac{\beta u}{\eta} ) }{u\sqrt{ ( \tau-s ) }}\exp \biggl( \frac{-\xi ^{2}}{4u}+q_{2} ( \tau-s ) -\eta_{0}u \biggr) \\ &{}\times J_{1} \bigl( 2\sqrt{\eta _{2}u ( \tau-s ) } \bigr) \, \mathrm{d}u \,\mathrm{d}s \\ &{}+\frac{\xi\sqrt{\eta_{1}\eta_{2}}}{2\sqrt{\pi}}\int_{0}^{\tau}\int _{0}^{\sigma}\int_{0}^{\infty} \frac{\delta ( \tau-s-\frac{\beta u}{\eta} ) }{\sqrt{u\sigma ( s-\sigma )}}\exp \biggl( \frac{-\xi^{2}}{4u}+q_{1} \sigma+q_{2} ( s-\sigma )-\eta_{0}u \biggr) \\ &{}\times J_{1} ( 2\sqrt{\eta_{1}u\sigma} ) J_{1} \bigl( 2\sqrt {\eta_{2}u ( s-\sigma ) } \bigr) \, \mathrm{d}u \,\mathrm{d}s \,\mathrm{d}\sigma. \end{aligned}$$
(19)
Introducing Eqs. (19) and (14) into Eq. (13), we get
$$\begin{aligned} G ( \xi,\tau ) =&\frac{\xi}{2\sqrt{\pi}}\int_{0}^{\tau } \int_{0}^{\infty}\frac{ ( \tau-s ) \delta ( s-\frac{\beta u}{\eta} ) }{u\sqrt{u}}\exp \biggl( \frac{-\xi^{2}}{4u}-\eta _{0}u \biggr) \,\mathrm{d}u \,\mathrm{d}s \\ &{}-\frac{\xi\sqrt{\eta_{1}}}{2\sqrt{\pi}}\int_{0}^{\tau }\int _{0}^{\sigma}\int_{0}^{\infty} \frac{ ( \tau -s ) \delta ( \sigma-\frac{\beta u}{\eta} ) }{u\sqrt{ s-\sigma}}\exp \biggl( \frac{-\xi^{2}}{4u}+q_{1} ( s-\sigma ) - \eta_{0}u \biggr) \\ &{}\times J_{1} \bigl( 2\sqrt{\eta_{1}u ( s-\sigma ) } \bigr) \,\mathrm{d}u \,\mathrm{d}s \,\mathrm{d}\sigma \\ &{}-\frac{\xi\sqrt{\eta_{2}}}{2\sqrt{\pi }}\int _{0}^{\tau}\int_{0}^{\sigma} \int_{0}^{\infty} \frac{ ( \tau-s ) \delta ( \sigma-\frac{\beta u}{\eta} ) }{u\sqrt{s-\sigma}}J_{1} \bigl( 2\sqrt{\eta_{2}u ( s-\sigma ) } \bigr) \\ &{}\times\exp \biggl( \frac{-\xi^{2}}{4u}+q_{2} ( s-\sigma ) - \eta_{0}u \biggr) \,\mathrm{d}u \,\mathrm{d}s \,\mathrm{d}\sigma \\ &{}+ \frac{\xi\sqrt{\eta _{1}\eta_{2}}}{2\sqrt{\pi}}\int_{0}^{\tau }\int _{0}^{\sigma}\int_{0}^{w} \int_{0}^{\infty }\frac{ ( \tau-s ) \delta ( s-\sigma-\frac{\beta u}{\eta } ) }{\sqrt{uw ( \sigma-w ) }} \\ &{}\times\exp \biggl( \frac{-\xi^{2}}{4u}+q_{1}w+q_{2} ( s- \sigma ) -\eta_{0}u \biggr) J_{1} ( 2\sqrt{ \eta_{1}uw} ) \\ &{}\times J_{1} \bigl( 2\sqrt{\eta_{2}u ( \sigma-w ) } \bigr) \,\mathrm{d}u \,\mathrm{d}s \,\mathrm{d}\sigma \,\mathrm{d}w. \end{aligned}$$
(20)
Taking \(u=\frac{v\eta}{\beta} \) into Eq. (20) and keeping in mind
$$ \int_{a}^{b}f ( x ) \delta ( x-x_{0} ) \,\mathrm{d}x=\left \{ \begin{array}{l@{\quad}l} f ( x_{0} )& \text{for }x\in [ a,b ), \\ 0&\text{for }x\notin [ a,b ) , \end{array} \right . $$
(21)
one finally obtains
$$\begin{aligned} G ( \xi,\tau ) =&\frac{\xi\sqrt{\beta}}{2\sqrt{\pi\eta}} \int_{0}^{\tau} \frac{ ( \tau-s ) }{s\sqrt{s}}\exp \biggl( \frac{-\beta\xi^{2}}{4\eta s}-\frac{\eta\eta_{0}}{\beta}s \biggr) \, \mathrm{d}s \\ &{}- \frac{\xi\sqrt{\eta_{1}}}{2\sqrt{\pi}}\int_{0}^{\tau }\int _{0}^{\sigma}\frac{ ( \tau-s ) }{\sigma\sqrt{ s-\sigma}}\exp \biggl( \frac{-\beta\xi^{2}}{4\eta\sigma}+q_{1} ( s-\sigma ) -\frac{\eta_{0}\eta}{\beta}\sigma \biggr) \\ &{}\times J_{1} \biggl( 2\sqrt{\frac{\eta_{1}\eta}{\beta}\sigma ( s-\sigma ) } \biggr)\, \mathrm{d}s \,\mathrm{d}\sigma \\ &{}- \frac{\xi\sqrt{\eta_{2}}}{2\sqrt{\pi}}\int_{0}^{\tau }\int _{0}^{\sigma}\frac{ ( \tau-s ) }{\sigma\sqrt{ s-\sigma}}\exp \biggl( \frac{-\beta\xi^{2}}{4\eta\sigma}+q_{2} ( s-\sigma ) -\frac{\eta_{0}\eta}{\beta}\sigma \biggr) \\ &{}\times J_{1} \biggl( 2\sqrt{\frac{\eta_{2}\eta}{\beta}\sigma ( s-\sigma ) } \biggr) \,\mathrm{d}s \,\mathrm{d}\sigma \\ &{}+ \frac{\xi\sqrt{\eta_{1}\eta_{2}\eta}}{2\sqrt{\beta\pi}} \int_{0}^{\tau}\int _{0}^{\sigma}\int_{0}^{w} \frac{ ( \tau-s ) }{\sqrt{w ( s-\sigma ) ( \sigma -w ) }} \\ &{}\times \exp \biggl( \frac{-\beta\xi^{2}}{4\eta ( s-\sigma )}+q_{1}w+q_{2} ( s-\sigma ) -\frac{\eta_{0}\eta}{\beta } (s-\sigma ) \biggr) \\ &{}\times J_{1} \biggl( 2\sqrt{\frac{\eta_{1}\eta}{\beta}w ( s-\sigma ) } \biggr) J_{1} \biggl( 2\sqrt{\frac{\eta_{2}\eta}{\beta} ( s-\sigma ) ( \sigma-w ) } \biggr) \,\mathrm{d}s \,\mathrm{d}\sigma \,\mathrm{d}w. \end{aligned}$$
(22)
The starting solution (22) holds for both small and large times. In order to write as a sum of the steady-state and transient solutions, we use the relation
$$ \int_{0}^{\tau}f ( \xi,\tau,s ) \, \mathrm{d} s=\int_{0}^{\infty}f ( \xi,\tau,s ) \, \mathrm{d} s-\int_{\tau}^{\infty}f ( \xi,\tau,s ) \, \mathrm{d}s. $$
(23)
Hence Eq. (22) reduces to the following form:
$$\begin{aligned} G ( \xi,\tau ) =&\tau\exp ( -\xi\sqrt{\eta _{0}} ) - \frac{\xi\beta}{2\eta\sqrt{\eta_{0}}}\exp ( -\xi\sqrt{\eta _{0}} ) \\ &{}-\frac{\xi\sqrt{\beta}}{2\sqrt{\pi\eta}}\int _{\tau }^{\infty}\frac{ ( \tau-s ) }{s\sqrt{s}}\exp \biggl( \frac {-\beta\xi^{2}}{4\eta s}-\frac{\eta\eta_{0}}{\beta}s \biggr) \,\mathrm{d}s \\ &{}- \frac{\xi\sqrt{\eta_{1}}}{2\sqrt{\pi}}\int_{0}^{\tau }\int _{0}^{\sigma}\frac{ ( \tau-s ) }{\sigma\sqrt{ s-\sigma}}\exp \biggl( \frac{-\beta\xi^{2}}{4\eta\sigma}+q_{1} ( s-\sigma ) -\frac{\eta_{0}\eta}{\beta}\sigma \biggr) \\ &{}\times J_{1} \biggl( 2\sqrt{\frac{\eta_{1}\eta}{\beta}\sigma ( s-\sigma ) } \biggr) \,\mathrm{d}s \,\mathrm{d}\sigma \\ &{}- \frac{\xi\sqrt{\eta_{2}}}{2\sqrt{\pi}}\int_{0}^{\tau }\int _{0}^{\sigma}\frac{ ( \tau-s ) }{\sigma\sqrt{ s-\sigma}}\exp \biggl( \frac{-\beta\xi^{2}}{4\eta\sigma}+q_{2} ( s-\sigma ) -\frac{\eta_{0}\eta}{\beta}\sigma \biggr) \\ &{}\times J_{1} \biggl( 2\sqrt{\frac{\eta_{2}\eta}{\beta}\sigma ( s-\sigma ) } \biggr) \,\mathrm{d}s \,\mathrm{d}\sigma \\ &{}+ \frac{\xi\sqrt{\eta_{1}\eta_{2}\eta}}{2\sqrt{\beta\pi}} \int_{0}^{\tau}\int _{0}^{\sigma}\int_{0}^{w} \frac{ ( \tau-s ) }{\sqrt{w ( s-\sigma ) ( \sigma -w ) }} \\ &{}\times \exp \biggl( \frac{-\beta\xi^{2}}{4\eta ( s-\sigma )}+q_{1}w+q_{2} ( s-\sigma ) -\frac{\eta_{0}\eta}{\beta } (s-\sigma ) \biggr) \\ &{}\times J_{1} \biggl( 2\sqrt{\frac{\eta_{1}\eta}{\beta}w ( s-\sigma ) } \biggr) J_{1} \biggl( 2\sqrt{\frac{\eta_{2}\eta}{\beta} ( s-\sigma ) ( \sigma-w ) } \biggr) \,\mathrm{d}s \,\mathrm{d}\sigma \,\mathrm{d}w. \end{aligned}$$
(24)